PAT 1020 Tree Traversals

发布时间：2020-12-14 04:32:07 所属栏目：大数据来源：网络整理

导读：1020?Tree Traversals?(25?分) ? Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences,you are supposed to output the level order traversal sequence of the correspondin

1020?Tree Traversals?(25?分)

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences,you are supposed to output the level order traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains one test case. For each case,the first line gives a positive integer?

Output Specification:

For each test case,print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space,and there must be no extra space at the end of the line.

Sample Input:

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

struct node{
    int data = -1;
    node *left = NULL;
    node *right = NULL;
};
int n,cnt;

int a[31];
int b[31];

node* createtree(int start,int end){
    if(start > end)
        return NULL;
    int num = a[cnt--];
    node *t = new(node);
    t->data = num;
    if(start == end){ return t;}

    int pos = -1;
    for(int i=start;i <= end;i++){
        if(b[i] == num) pos = i;
    }

    t->right = createtree(pos+1,end);
    t->left = createtree(start,pos-1);

    return t;
}



int main(){
    cin >> n;
    cnt = n-1;

    for(int i=0;i < n;i++) cin >> a[i];
    for(int i=0;i < n;i++) cin >> b[i];

    node *root = createtree(0,n-1);

    queue<node*> que;
    que.push(root);
    int count = 0;

    while(!que.empty()){
        node *t = que.front();que.pop();
        cout << t->data;
        if(count!=n-1){cout << " ";count++;}
        if(t->left)que.push(t->left);
        if(t->right)que.push(t->right);
    }



    return 0;
}

创建节点要new，不然就会出错

（编辑：李大同）

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